Warm-up Question: 1

• What was the most confusing part of Monday’s

lecture?
A. The sign of shear stress
B. Why do we need plane stress?
C. Wait, what is plane stress?
D. Mohr’s circle
E. Principal stresses and max shear stress
 7

http://www.gereports.com/post/112705004705/ge-develops-jet-engines-made-from-ceramic-matrix/
 8

http://www.mece.ualberta.ca/~wmoussa/lab/research.
htm

http://galleryhip.com/tensile-test.html
 Objectives 9

• Use the condition of triaxial stress to determine the

maxiumum shear stress from principle stresses
• Define the condition of plane strain
• Perform coordinate transformation of plane strain
• Use Mohr’s circle to visualize coordinate transformations of
strain and find the maximum shear strain
 Mohr s circle 10

• Mohr s circle is a convenient way to visually remember the two-

dimensional transformation equations.
• Approach:
1.Make 2D axes :
• right = positive normal stress
• down = positive shear stress
2.Plot (σx, τxy) and (σy, −τxy) (remember sign convention for τ!)
3.Draw a line between both points to be diameter of circle.
4.Draw circle centered at midpoint of line.
5.Rotate by points by 2θ, read off new (σx’, τx’y’) and (σy’, −τx’y’)
 Mohr’s circle resources 11

• https://www.youtube.com/watch?v=-bIh3xSZBqE
• http://www.ijee.ie/OnlinePapers/Interactive/Philpot/mo
hr_learning_tool.htm
 Mohr s circle 12

max normal
stress

min normal
stress ave normal
stress

max shear
stress
Example 13
 Question (14): 14

• Due to the applied loading, the element at point A on the solid shaft is
subjected to the state of stress shown. Determine the principal stresses
acting at this point.
First: What is the stress state?
A. σx = −12 ksi, σy = 0 ksi, τxy = +6 ksi
B. σx = −12 ksi, σy = 0 ksi, τxy = −6 ksi
C. σx = +12 ksi, σy = 0 ksi, τxy = +6 ksi
D. σx = +12 ksi, σy = 0 ksi, τxy = −6 ksi
 Question (15): 15

• Due to the applied loading, the element at point A on the solid shaft is
subjected to the state of stress shown. Determine the principal stresses
acting at this point.
Next: Where is the center for Mohr s circle?
A. (0 ksi, 0 ksi)
B. (−6 ksi, 0 ksi)
C. (−12 ksi, 0 ksi)
D. (−6 ksi, −6 ksi)
E. (−12 ksi, −6 ksi)
 Question (16): 16

• Due to the applied loading, the element at point A on the solid shaft is
subjected to the state of stress shown. Determine the principal stresses
acting at this point.
Next: What is the radius for Mohr s circle?
A. 6.00 ksi
B. 8.49 ksi
C. 12.0 ksi
D. 19.0 ksi
 Question (17): 17

• Due to the applied loading, the element at point A on the solid shaft is
subjected to the state of stress shown. Determine the principal stresses
acting at this point.
Finally: What are the principal stresses?
A. −1.76 ksi, −10.2 ksi
B. 0 ksi, −12.0 ksi
C. 2.49 ksi, −14.5 ksi
D. 11.9 ksi, −23.0 ksi
 Example 18

• Due to the applied loading, the element at point A on the solid shaft is
subjected to the state of stress shown. Determine the principal stresses
acting at this point.
 Transformation of stress and strain 21

Plane stress
Simplification of loading to a plane
Sign convention
Transformation of coordinates
Transforming loads and areas from one
reference frame (coordinate axes) to
another
Principal stresses and directions
Eigenvalues and eigenvectors of stress
Max/min normal stress
Max shear stress
Mohr s circle
geometric device to compute transform
Multiple loads in 3D
principle stresses, max shear stress
Plane strain
simplify a deformation to a single plane
Transformation of coordinates
transforming displacements from one
reference frame to another
 General state of stress 22

• A general state of stress has 6 components:

• 3 normal: σx σy σz
• 3 shear: τyz τxz τxy
• The components depend on the choice of
coordinate system: xyz
• For a different cube section, equilibrium can
be applied to faces as before.
• Principal stresses in 3D are defined the
same as before:
• Identify three ________________that
have only ______________
• Normal stress: σA is pointed along A
(where A is the vector normal for a plane)
• Principal stresses/planes =
eigenvalue/eigenvectors for stress.
 Graphical stress tensor components 23

• Stress area = force Fi = Σj=xyz σij Aj

σxx=σx σxy=τxy σxz=τxz

σyx=τxy σyy=σy σyz=τyz

σzx=τxz σzy=τyz σzz=σz

 Triaxial stress 24

• Principal stresses in 3D
• Solutions to eigenvalue problem:
• σA = sA
• Three perpendicular A s each with own
eigenvalue s
• Sort eigenvalues (principal stresses):
σmin ≤ σint ≤ σmax
• For plane stress...
• One principal stress is _____
(perpendicular to plane)
• Other two are principal stresses in plane
• ______________ into σmin ≤ σint ≤ σmax
depending on _____ of in-plane principal
stresses
 Maximum shear stress 25

• Given three principal stresses

σmin ≤ σint ≤ σmax
we can find the maximum shear stresses in
the three planes:
1.min-int: τ = (σint − σmin)/2
2.min-max: τ = (σmax − σmin)/2
3.int-max: τ = (σmax − σint)/2

τ = (σint − σmin)/2

τ = (σmax − σint)/2

τ = (σmax − σmin)/2
 Maximum shear stress: plane stress 26

• The same ideas apply for plane stress,

where one principal stress is zero.
• Given three principal stresses
σmin ≤ σint ≤ σmax
we can find the maximum shear stresses in
the three planes:
1.min-int: τ = (σint − σmin)/2
2.min-max: τ = (σmax − σmin)/2
3.int-max: τ = (σmax − σint)/2

0 ≤ σ2 ≤ σ1 σ2 ≤ 0 ≤ σ1
 Question: 27

• The point on the surface of the cylindrical pressure vessel is subjected to

a state of plane stress. Determine the absolute maximum shear stress at
this point.
First: What are the three principal
stresses?
A. −32 MPa, −16 MPa, 0 MPa
B. −16 MPa, 0 MPa, +32 MPa
C. −16 MPa, +8 MPa, +32 MPa
D. 0 MPa, +16 MPa, +32 MPa
E. +16 MPa, +24 MPa, +32 MPa
 Question: 28

• The point on the surface of the cylindrical pressure vessel is subjected to

a state of plane stress. Determine the absolute maximum shear stress at
this point.
Next: What is the absolute maximum
shear stress?
A. 0 MPa
B. 8 MPa
C. 16 MPa
D. 24 MPa
E. 32 MPa
 Question: 29

• The point on the surface of the cylindrical pressure vessel is subjected to

a state of plane stress. Determine the absolute maximum shear stress at
this point.
Finally: What is the maximum
shear stress in the plane of loading?
A. 0 MPa
B. 8 MPa
C. 16 MPa
D. 24 MPa
E. 32 MPa
 Example 30

• The point on the surface of the cylindrical pressure vessel is subjected to

a state of plane stress. Determine the absolute maximum shear stress at
this point.
 Plane strain 37

• A general state of strain has 6 components:

• 3 normal: ϵx ϵy ϵz
• 3 shear: γyz γxz γxy
• The components depend on the choice of
coordinate system: xyz
• There may be a plane which contains the
only nonzero strain components
• E.g., all deformations contained in a plane
• Choose the xy coordinate system to
contain the nonzero strains
• 2 normal: ϵx ϵy
• 1 shear: γxy
• The components depend on the choice of
coordinate system: xy
• Note: ____________ ≠ _____________
• Due to Poisson effect (next week!):
normal stresses in xy plane create
normal strains in z direction, too.

plane stress ≠ plane strain

 Transforming plane strain 38

• A state of plane strain has three components

• The components are _______________ of the
________________________ making it into a parallelogram
• ϵx, ϵy, γxy all depend on how that square is oriented.
• The components for one choice can be determined from the
components for another choice by constructing ________, and taking
___________.
• Remember the definitions of strain: change in a vector
• ϵx = change of i in direction of i
• ϵy = change of j in direction of j
• γxy = half change of i in direction of j, half change of j in direction of i
before after
strain strain
 Graphical strain tensor components 39

• Strain length = length change δℓi = Σj=xyz εij ℓj

εxx=ϵx εxy=½γxy εxz=½γxz

εyx=½γxy εyy=ϵy εyz=½γyz

εzx=½γxz εzy=½γyz εzz=ϵz

 Transforming plane strain 40

• Remember the definitions of strain: change in a vector

• ϵx = change of i in direction of i
• ϵy = change of j in direction of j
• γxy = half change of i in direction of j, half change of j in direction of i
• Consider ϵx : change of i in direction of i
 Transforming plane strain 41

• Remember the definitions of strain: change in a vector

• ϵx = change of i in direction of i
• ϵy = change of j in direction of j
• γxy = half change of i in direction of j, half change of j in direction of i
• Consider ϵx : change of i in direction of i

dot product
with i’
 Transforming plane strain 42

• Remember the definitions of strain: change in a vector

• ϵx = change of i in direction of i
• ϵy = change of j in direction of j
• γxy = half change of i in direction of j, half change of j in direction of i
• Consider γx y : change of i in direction of j

dot product
with j’
 Transforming plane strain 44

• Remember the definitions of strain: change in a vector

• ϵx = change of i in direction of i
• ϵy = change of j in direction of j
• γxy = half change of i in direction of j, half change of j in direction of i
• Consider ϵy : change of j in direction of j
 Transforming plane strain 46

• Remember the definitions of strain: change in a vector

• ϵx = change of i in direction of i
• ϵy = change of j in direction of j
• γxy = half change of i in direction of j, half change of j in direction of i
• Consider ϵy : change of j in direction of j

dot product
with j’
 Transforming plane strain: summary 48

• Transforming strain is the same as for stress: σ→ϵ, τ→½γ

• We can simplify the expressions by using double-angle formulae
• sin 2θ = 2 sin θ cos θ
• cos 2θ = cos2θ − sin2θ

• Note: ϵx + ϵy = ϵx + ϵy
• Note the similarity in equations—what we did for stress, we do for strain:
• Same special cases: 180 , 90 , 45 rotations
• Mohr s circle to rotate in plane
• Principal strains (max/min normal strain) =
• Max shear strain
• Be mindful of the 1/2 factor for shear in Mohr s circle!
 Question: 49

• A differential element of material at a point is subjected to a state of

plane strain ϵx=500(10−6), ϵy=−300(10−6), γxy=200(10−6). Determine the
equivalent strains acting on an element of material oriented at the point
clockwise 30 from the original position.
First: What is the average normal strain?
A. +400(10−6)
B. +200(10−6)
C. +100(10−6)
D. −100(10−6)
E. −200(10−6)
 Question: 50

• A differential element of material at a point is subjected to a state of

plane strain ϵx=500(10−6), ϵy=−300(10−6), γxy=200(10−6). Determine the
equivalent strains acting on an element of material oriented at the point
clockwise 30 from the original position.
Next: What are cos 2θ and sin 2θ?
A. cos 2θ=+√3/2, sin 2θ=+1/2
B. cos 2θ=+1/2, sin 2θ=+√3/2
C. cos 2θ=+√3/2, sin 2θ=−1/2
D. cos 2θ=+1/2, sin 2θ=−√3/2
 Question: 51

• A differential element of material at a point is subjected to a state of

plane strain ϵx=500(10−6), ϵy=−300(10−6), γxy=200(10−6). Determine the
equivalent strains acting on an element of material oriented at the point
clockwise 30 from the original position.
Next: What is ϵx ?
A. +386(10−6)
B. +213(10−6)
C. −13.4(10−6)
D. −186(10−6)
 Question: 52

• A differential element of material at a point is subjected to a state of

plane strain ϵx=500(10−6), ϵy=−300(10−6), γxy=200(10−6). Determine the
equivalent strains acting on an element of material oriented at the point
clockwise 30 from the original position.
Next: What is ϵy ?
A. +386(10−6)
B. +213(10−6)
C. −13.4(10−6)
D. −186(10−6)
 Question: 53

• A differential element of material at a point is subjected to a state of

plane strain ϵx=500(10−6), ϵy=−300(10−6), γxy=200(10−6). Determine the
equivalent strains acting on an element of material oriented at the point
clockwise 30 from the original position.
Finally: What is γx y ?
A. +793(10−6)
B. +396(10−6)
C. −296(10−6)
D. −592(10−6)
 Example 54

• A differential element of material at a point is subjected to a state of

plane strain ϵx=500(10−6), ϵy=−300(10−6), γxy=200(10−6). Determine the
equivalent strains acting on an element of material oriented at the point
clockwise 30 from the original position.

+793(10−6) / 2
−13.4(10−6)

+793(10−6) / 2

+213(10−6)
 Principal strains 56

• The magnitude of normal and shear strains depends on orientation of

section with respect to the deformation
• From a given plane strain, we can find the angle that produces an
extremum in normal strain (max or min):

• Produces two solutions: θp1 and θp2, which are 90 apart (max and min)
• Finding the max and min normal stresses by substituting θp1 or θp2:

• Orientation for max shear strain found similarly:

• In that coordinate system:

 Principal strains 57

• Principal strains are the __________________________normal strains.

• The ______ on which these act are the principal _____________.
• Note: if a material has ___________ mechanical response then the
principal directions are the _____________ for the principal planes.
• This is because in an isotropic material, normal stresses only produce
normal strains, never shear strains.
• The principal directions experience no shear strain
• Using the principal directions as coordinate axes, there are no off-
diagonal components to the strain matrix.
• The strain tensor is diagonal, which makes the principal strains the
eigenvalues of the strain tensor, and the principal directions the
eigenvectors of the strain tensor.
 Mohr s circle for strain 58

• Mohr s circle is a convenient way to visually remember the two-

dimensional transformation equations.
• Approach:
1.Make 2D axes :
• right = positive normal strain
• down = positive shear strain divided by 2
2.Plot (ϵx, ½γxy) and (ϵy, −½γxy) (remember sign convention for γ/2!)
3.Draw a line between both points to be diameter of circle.
4.Draw circle centered at midpoint of line.
5.Rotate by points by 2θ, read off new (ϵx’, ½γx’y’) and (ϵy’, −½γx’y’)
 Mohr s circle for strain 59

ave normal max normal

strain strain

min normal
strain

max shear
strain/2
 Question: 60

• A differential element of material at a point is subjected to a state of

plane strain ϵx=−350(10−6), ϵy=200(10−6), γxy=80(10−6). Determine the
principal strains at the point and the orientation of principal directions.
First: What is the average normal strain?
A. −75.0(10−6)
B. −150(10−6)
C. −275(10−6)
D. −550(10−6)
 Question: 61

• A differential element of material at a point is subjected to a state of

plane strain ϵx=−350(10−6), ϵy=200(10−6), γxy=80(10−6). Determine the
principal strains at the point and the orientation of principal directions.
Next: What is the radius of Mohr s circle?
A. 85.0(10−6)
B. 110(10−6)
C. 278(10−6)
D. 286(10−6)
 Question: 62

• A differential element of material at a point is subjected to a state of

plane strain ϵx=−350(10−6), ϵy=200(10−6), γxy=80(10−6). Determine the
principal strains at the point and the orientation of principal directions.
Next: What are the principal strains?
A. −353(10−6), +203(10−6)
B. −203(10−6), +353(10−6)
C. −214(10−6), +64.0(10−6)
D. −64.0(10−6), +214(10−6)
 Question: 63

• A differential element of material at a point is subjected to a state of

plane strain ϵx=−350(10−6), ϵy=200(10−6), γxy=80(10−6). Determine the
principal strains at the point and the orientation of principal directions.
Next: What are the principal directions
relative to x?
A. −85.9 and +4.14
B. −8.28 and +172
C. −4.14 and +85.9
D. −172 and +8.28
 Question: 64

• A differential element of material at a point is subjected to a state of

plane strain ϵx=−350(10−6), ϵy=200(10−6), γxy=80(10−6). Determine the
principal strains at the point and the orientation of principal directions.
Finally: What is the maximum shear strain?
A. 85.0(10−6)
B. 110(10−6)
C. 278(10−6)
D. 286(10−6)
E. 556(10−6)
 Example 65

• A differential element of material at a point is subjected to a state of

plane strain ϵx=−350(10−6), ϵy=200(10−6), γxy=80(10−6). Determine the
principal strains at the point and the orientation of principal directions.

+203(10−6)

−353(10−6)
 Summary 67

• Used the condition of triaxial stress to determine the maxiumum shear

stress from principle stresses
• Defined the condition of plane strain
• Performed coordinate transformation of plane strain
• Used Mohr’s circle to visualize coordinate transformations of strain and
find the maximum shear strain